Question

How do i do a and b?

A package of mass m is released from rest at a warehouse loading dock and slides down the h= 3.2 m- high, frictionless chute to a waiting truck. Unfortunately, the truck driver went on a break without having removed the previous package, of mass 2m, from the bottom of the chute. (Figure 1) Suppose the packages stick together. What is their common speed after the collision? Express your answer to two significant figures and include the appropriate units. Suppose the collision between the packages is perfectly elastic. To what height does the package of mass m rebound? Express your answer to two significant figures and include the appropriate units.

Answer 1

Given that,

height at which mass m is present h = 3.2 meters

mass of m1 = m and that of m2 = 2m

The gravitational potential energy of m gets convereted to kinetic energy of motion, so for m we can write:

PE = KE

m g h = 1/2 m v1i² (m gets cancelled)

v1i = sqrt(2 g h) = sqrt (2 x 9.8 x 3.2) = 7.92 m/s = 8 m/s

(a)The intial momentum of the given system is due to mass m since 2m is resting intiallyLet V be the speed after they sticks together. So, from the conservation of linear momentum we have:

Pi = Pf

m v1i = (m1 + m2)V

V = mv1i / (m1 + m2) = m x 8m/s / (3 m ) = 2.64 m/s

Hence, after they stick together, common speed = V = 2.64 m/s

(b)we have the initial velocity of mass m before it collised with mass 2m. We need to know the velocity with which the mass m rebounds back to the ramp to know the height to which it rebounds back. Let it be v1f.

If the collision is elastic so the velocity v1f will be given by:

v1f =( m1-m2 )v1i/(m1+m2)

v1f = m - 2m x 7.92 / (m +2m) = -m/3m x 7.92 = - 2.64 m/s

Now again using comservation of energy for m1, after it collides with m2 and bounds back to the ramp. Let H be the height to which it bounds back, so

1/2 m1 v1f² = m1 g H (m1 gets cancelled)

H = v1f² / 2 g = (-2.64)²/2 x 9.81 = 0.36 meters

Hence, height at which it rebounds back = H = 0.36 meters.